document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=860312>Question 860312</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Does 3 divide(3k+1)(3k+2)(3k+3)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #518416 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=518416\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> note for n, d elements of Z and d is not = 0, then n is divisible by d if and only if, n = d * k for some k in Z.<BR>\n" );
document.write( "now<BR>\n" );
document.write( "Does 3 divide(3k+1)(3k+2)(3k+3)?  YES<BR>\n" );
document.write( "Proof<BR>\n" );
document.write( "(3k + 1)(3k + 2)(3k + 3) = 3(3k + 1)(3k + 2)(k + 1)<BR>\n" );
document.write( "since k is an element of Z, it follows that (3k + 1)(3k + 2)(k + 1) is an element in Z, this and our definition of divisible gives us the desired result.<BR>\n" );
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